If the value of $\int {x\sin x = - x cosx + \alpha } $ , then the value of $\alpha $ is :
A. $\sin x$ +C
B. $\cos x$ +C
C. C
D. None of these

Answer Verified Verified
Hint: Whenever there is a product of the two functions, then integration by parts or partial integration should be used. For instance, if and are the function of x, the integration of \[\int {u.vdx} = u\int {vdx} - \int {\left[ {\left( {\dfrac{{du}}{{dx}}} \right)\int {vdx} } \right]} dx\] , where $u = $ first function and $v = $ second function.

Complete step by step solution:$\int {x\sin x = - x cosx + \alpha } ......(1)$
From the integration it is clear that it is a product of two functions of x. Therefore, integration by parts should be used to evaluate the integral.
Let us assume x as the first function and as the second function.
\[I = x\int {\sin xdx} - \int {\left[ {\left( {\dfrac{{d\left( x \right)}}{{dx}}} \right)\int {\sin xdx} } \right]dx} ......(2)\]
The integration of $\int {\sin xdx = - co} sx$ and the differentiation of $x = 1$ , put the value in equation (2)
$I = - x\cos x + \int {\cos xdx} ......(3)$
The integration of \[\int {\cos xdx = \sin } x\] put the value in equation (3)
$I = - x\cos x + \sin x + C......(4)$
Where, C is a constant and known as constant of integration.
Comparing equation (4) and (1), it is clear that
The value of $\alpha = \sin x + C$
Hence, the correct option is (A).

Note: The choice of the first and second function should be in accordance with the thumb rule, which says that choose the function in the order of ILATE where I is the inverse trigonometric function, L is the logarithmic function, A is the algebraic function, T is the Trigonometric function and E is the exponential function.
For instance in $\int {x\cos x} $ x is the first function and is the second function because x is the algebraic function and cosine function is the trigonometric function. So the preference for first function is given to x.